Two bodies of masses m_1 = 200 g and m_2 = 300 g hang from…

Two bodies of masses $m_1 = 200\ \mathrm{g}$ and $m_2 = 300\ \mathrm{g}$ hang from the ends of a rope passing over a frictionless pulley of negligible mass. Find the acceleration of the system.

Explanation of Method

Atwood machine with 200 g and 300 g.

Step 1: Equations of motion

The heavier body $m_2$ goes down, $m_1$ goes up, both with acceleration $a$: $T - m_1 g = m_1 a$.

\[ m_2 g - T = m_2 a. \]

Step 2: Acceleration

\[ a = \frac{m_2 - m_1}{m_1 + m_2}\,g = \frac{0.1}{0.5} \cdot 9.81. \]

\[a \approx 1.96\ \mathrm{m/s^2}\]

The rope tension is $T = m_1(g + a) \approx 2.35\ \mathrm{N}$.

$0.2\ \mathrm{kg}$ $0.3\ \mathrm{kg}$ $a$ $a$ $a = \frac{m_2 - m_1}{m_1 + m_2}\,g$

Atwood machine. Only the difference of the weights, $(m_2 - m_1)g$, drives the motion, but it has to accelerate both masses.

Result

\[\boxed{\,a \approx 1.96\ \mathrm{m/s^2}\,}\]

Remark

Check. The original result is correct: $a = g/5$.